Question

我在C中编写了一个程序，用于计算函数exp（x）在x的各种值的近似值，有些是正数，有些是负数。我的问题如下：

尽管之前我能够在序列中包含151个术语，现在，即使在第140届，该系列也会爆发exp（100）。之后我用代码搞砸了更多，现在系列爆炸了甚至在130学期。除此之外我还用了很长时间数字格式加倍，以允许大的阶乘计算和显示。怎么能解释这个现象呢？有时连续执行代码后我得到了结果近似我检查的函数的实际值 Wolfram Alpha。
我听说“for”循环被证明效率低下;你有没有记住“for”循环的替代方案？我很乐意尝试尽可能多地实施它们。
我想以科学格式显示输出但格式说明者（我不确定这个术语）不允许我这样做。我在printf语句中使用的说明符是％g。我怎么能拥有输出以科学格式显示？
根据我使用的硬件和软件，这是最大的我能得到什么？
我还得到了分段错误（核心转储）错误而不是显示x = -100。这背后可能是什么原因？

感谢您的贡献。我感谢任何帮助和建议。

P.S：我在64位硬件上使用GCC编译，但在32位Ubuntu 12.04 LTS（a.k.a Precise Pangolin）上编译。

#include <stdio.h>
#include <math.h>
//We need to write a factorial function beforehand, since we
//have factorial in the denominators.
//Remembering that factorials are defined for integers; it is
//possible to define factorials of non-integer numbers using
//Gamma Function but we will omit that.
//We first declare the factorial function as follows:
long double factorial (double);
//Long long integer format only allows numbers in the order of 10^18 so 
//we shall use the sign bit in order to increase our range.
//Now we define it,
long double
factorial(double n)
{
//Here s is the free parameter which is increased by one in each step and
//pro is the initial product and by setting pro to be 0 we also cover the
//case of zero factorial.
    int s = 1;
    long double pro = 1;
    if (n < 0)
        printf("Factorial is not defined for a negative number \n");
    else {
    while (n >= s) { 
    pro *= s;
    s++;
    }
    return pro;
    }
}

int main ()
{
    long double x[13] = { 1, 5, 10, 15, 20, 50, 100, -1, -5, -10, -20, -50, -100};
//Here an array named "calc" is defined to store 
//the values of x.
//The upper index controls the accuracy of the Taylor Series, so
//it is suitable to make it an adjustable parameter. However, after
//a certain value of p the series become infinitely large to be 
//represented by the computer; hence we terminate the series at 
//the 151th term. 
int p = 150;
long double series[13][p];
int i, k;
//We only define the Taylor series for positive values of the exponent, and
//later we will use these values to calculate the reciprocals. This is done
//in this manner to avoid the ambiguity introduced into the sum due to terms
//alternating in sign.
long double sum[6] = { 0 };
for (i = 0; i <= 6;i++) {

        for (k = 0; k <= p; k++){
        series[i][k] = pow(x[i], k)/( factorial(k));
        sum[i] += series[i][k];
}
printf("Approximation for x = %Lf is %Lf \n", x[i], sum[i]);
}
//For negative numbers -taking into account first negative number is
//in the 8th position in the x array and still denoting the approximation
long double approx[5] = { 0 };
for (i = 7; i <= 12;i++) {
    approx[i - 7] = 1 / sum[i - 7];
    printf("Approximation for x = %Lf is %Lf \n", x[i], approx[i - 7]);
}
//printf("%Lf \n", factorial(3));
//The above line was introduced to test if the factorial function
//was functioning properly.
}

Answer 1

回答你的一些问题。

1我的日食代码没有爆炸。

2除了可能过多的迭代之外，我认为使用for循环没有固有的低效率。

3使用"%Le"作为科学格式。

4您可以获得的最大精度将是LDBL_EPSILON 1.取决于您不知道的系统（可能是1个功率（2,64））。记住精度是相对的，而不是绝对的。

5分段错误可能是由于approx[5]造成的。 @Grijesh Chauhan

建议

将-15添加到long double x[13+1] = { 1, 5, 10, 15, 20, 50, 100, -1, -5, -10, -15, -20, -50, -100 };。更改其他各种13 s。

添加返回错误条件：
if (n < 0) { printf("Factorial is not defined for a negative number \n"); return 0; }

更改为long double approx[5+1+1] = { 0 };

添加返回main()。

如果您首先添加小条款，则会获得更好的数值结果。这通常意味着撤消for (k = 0; k <= p; k++)循环。

int main() {
  long double x[13+1] = { 1, 5, 10, 15, 20, 50, 100, -1, -5, -10, -15, -20, -50, -100 };
  int p = 150;
  long double series[13+1][p];
  int i, k;

  long double sum[6+1] = { 0 };
  for (i = 0; i <= 6; i++) {
    for (k = 0; k <= p; k++) {
      series[i][k] = pow(x[i], k) / (factorial(k));
      sum[i] += series[i][k];
    }
    printf("Approximation for x = %Le is %Le %e\n", x[i], sum[i], exp(x[i]));
  }

  long double approx[7] = { 0 };
  for (i = 7; i <= 12; i++) {
    approx[i - 7] = 1 / sum[i - 7];
    printf("Approximation for x = %Le is %Le %e\n", x[i], approx[i - 7], exp(x[i]));
  }
  return 0;
}

用GCC编译的C系列泰勒级数计算程序

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